For positive numbers x,y and z numerical value of determinant left| {,begin{array}{*{20}{c}}1

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

easy

For positive numbers $x,y$ and $z$ the numerical value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$is

$0$

$1$

${\log _e}xyz$

None of these

(IIT-1993)

Solution

(a) $\left| {\,\begin{array}{*{20}{c}}1&{{{\log }_x}y}&{{{\log }_x}z}\\{{{\log }_y}x}&1&{{{\log }_y}z}\\{{{\log }_z}x}&{{{\log }_z}y}&1\end{array}\,} \right|$

= $(1 – {\log _z}y{\log _y}z) – {\log _x}y({\log _y}x – {\log _z}x{\log _y}z)$$ + {\log _x}z({\log _y}x{\log _z}y – {\log _z}x)$

= $(1 – 1)\, – (1 – {\log _x}y{\log _y}x) + ({\log _x}z{\log _z}x – 1) = 0$ $\{$ Since ${\log _x}y.{\log _y}x = 1\} $

Standard 12

Mathematics

Find equation of line joining $(1,2)$ and $(3,6)$ using determinates

easy

If $'a'$ is non real complex number for which system of equations $ax -a^2y + a^3z$ = $0$ , $-a^2x + a^3y + az$ = $0$ and $a^3x + ay -a^2z$ = $0$ has non trivial solutions, then $|a|$ is

normal

The remainder when the determinant $\left|\begin{array}{lll} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{array}\right|$ is divided by $5$ is

normal

(KVPY-2015)

The determinant $\left| {\,\begin{array}{*{20}{c}}{4 + {x^2}}&{ – 6}&{ – 2}\\{ – 6}&{9 + {x^2}}&3\\{ – 2}&3&{1 + {x^2}}\end{array}\,} \right|$ is not divisible by

medium

The solutions of the equation $\left| {\,\begin{array}{*{20}{c}}x&2&{ – 1}\\2&5&x\\{ – 1}&2&x\end{array}\,} \right| = 0$ are